\(PTK_{M_2O}=2NTK_M+16=62\\ \Rightarrow NTK_M=23(đvC)\)

Vậy M là natri (Na)

CTHH: M₂O---> M₂=62 -16=46:2=23

=> M là nguyên tử Natri

\(C\%\left(H_3PO_4\right)=\dfrac{m_{H_3PO_4}}{m_{H_3PO_4}+200}.100\%=3,81\%\)

\(\Rightarrow m_{H_3PO_4}=7,92\left(g\right)\)

\(\Rightarrow n_{H_3PO_4}=0,08\left(mol\right)\)

\(\Rightarrow n_{P_2O_5}=0,04\left(mol\right)\)

\(\Rightarrow m=1,24\)

\(C\%\left(H_3PO_4\right)=\frac{m_{H_3PO_4}}{m_{H_3PO_4}+200}.100\%=3,81\%\)

\(\Rightarrow m_{H_3PO_4}=7,92g\)

\(\Rightarrow n_{H_3PO_4}=0,08mol\)

\(n_{P_2O_5}=0,04mol\)

\(\Rightarrow m=1,24\)

\(n_{Zn}=\dfrac{13}{65}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{HCl}=0,4(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,4.36,5}{100}.100\%=14,6\%\\ c,n_{ZnCl_2}=n_{H_2}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ \Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{13+100-0,2.2}.100\%\approx 24,16\%\)

Ủa thịnh THCS HTA đúng kh? Này đề thi giữa kì hóa mà thịnh đi hỏi hả ;) tui méc thầy nha

5

3

\(n_{H_2}=\dfrac{12,395}{24,79}=0,5(mol)\\ n_{O_2}=\dfrac{6,1975}{24,79}=0,25(mol)\\ \Rightarrow m_{hh}=m_{O_2}+m_{H_2}=0,5.2+0,25.32=1+8=9(g)\)

\(n_{H_2}=\dfrac{12,395}{22,4}=0,55mol\)

\(n_{O_2}=\dfrac{6,1975}{22,4}=0,28mol\)

\(m_{hh\uparrow}=m_{H_2}+m_{O_2}=0,55\cdot2+0,28\cdot32=13,26g\)

\(3Ag+4HNO_3\rightarrow3AgNO_3+NO+2H_2O\)

\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

\(NO\) là sản phẩm khử duy nhất.

\(\Rightarrow n_{NO}=\dfrac{0,448}{22,4}=0,02mol\)

Ta có: \(\left\{{}\begin{matrix}27n_{Al}+108n_{Ag}=3,51g\\BTe:3n_{Al}+n_{Ag}=3n_{NO}=0,06\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,01mol\\n_{Ag}=0,03mol\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,01\cdot27}{3,51}\cdot100\%=7,7\%\)

\(\%m_{Ag}=100\%-7,7\%=92,3\%\)

a, \(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

\(3Ag+4HNO_3\rightarrow3AgNO_3+NO+2H_2O\)

Câu 6:

\(a,m_{N_2O}=\dfrac{10,08.44}{22,4}=19,8(g)\\ m_{CO_2}=\dfrac{44.0,12.10^{-23}}{6.10^{-23}}=0,88(g)\\ b,\overline{M}=\dfrac{19,8+0,88}{0,45+0,02}=44(g/mol)\\ c,d_{X/kk}=\dfrac{44}{29}\approx 1,52\)

Vậy X nặng hơn kk khoảng 1,52 lần

Câu 5.

\(d\)_A/kk\(=\dfrac{64}{29}\)\(\Rightarrow M_A=\dfrac{64}{29}\cdot29=64đvC\)